For a SciFi project I work on I have a space-station orbiting earth that uses an 8 hour 3 shift system for its staff. Each shift is assigned to one-third of the countries and organizations on earth - namely those that are facing the station during the ~8 hours of the shift.

The above premise requires the station to be placed so that it will stay relatively put in regard to the earth and one of her full rotations.

Q: Where can I put my station so it can observe a full rotation of earth?

  • 2
    $\begingroup$ I would recommend editing the title of this question. The title implies to me (and, judging from the answers provided so far, others) that you are asking about either geosynchronous or geostationary orbit (the latter being my first assumption). The body of your question is phrased well, though; I fully understand what you are asking, despite it apparently being something different from what the title indicates. $\endgroup$
    – DanDoubleL
    Commented Sep 14, 2017 at 14:04
  • $\begingroup$ @DaniellYancey please feel free to propose a clearer title :) I tried it twice so far with only moderate success :/ $\endgroup$
    – dot_Sp0T
    Commented Sep 14, 2017 at 14:06
  • $\begingroup$ I'd say that the edit made by @KareemElashmawy is a rather good title. The use of "observes" helps with clarity. $\endgroup$
    – DanDoubleL
    Commented Sep 14, 2017 at 16:24
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    $\begingroup$ It's a common misconception I see often with astronomy related questions. People tend to forget that frames of reference are much more apparent in space. $\endgroup$ Commented Sep 14, 2017 at 17:01
  • $\begingroup$ Do the station staffers actually need line of sight to the people on the ground? You could always put the station somewhere mechanically convenient and use signal relay satellites to keep in touch with the people on the ground, wherever they are. It wouldn't necessarily matter where the station is physically located in space, if the staffers were all scheduled based on Zulu time or GMT or something. $\endgroup$ Commented Sep 14, 2017 at 17:25

10 Answers 10


From what I understand your requirements are thus:

  1. The station uses a 24 hour time-keeping system split into 3 shifts, 8 hours each.
  2. Each shift is staffed by members of the countries facing the station during that shift. Typically this amounts to 1/3 of Earth's organizations.
  3. The station must orbit Earth in such a manner that it observes Earth's entire rotation. You expect that it must be stationary for this.


Lagrange Points.

In astronomy and orbital mechanics, Lagrange points are positions in an orbital configuration where an object remains fixed relative to the other bodies. Point 3 automatically removes typical orbital positions such as Low-Earth-Orbit and Geosynchronous-Earth-Orbits. This leaves the Lagrange points as a natural selection; but, there are 4 such points for Earth and The Moon and 4 more for Earth and The Sun.

Earth-Moon Lagrange Points

If we look at the Earth-moon system, L1 is a natural point to choose for the station. It's situated between Earth and The Moon with relatively quick access to either body and can observe Earth's full rotation with ease. That being said, L1 poses 2 issues:

  1. L1 is a unstable point. If the satellite drifts a bit, it can fall out of L1 and towards the Earth or Moon in an unstable elliptical orbit. Unfortunately this applies to all Lagrange points, but unequally. Depending on your SciFi scenario, this may not be an issue assuming the station has basic thrust capacity to keep it within the Lagrange point.
  2. L1 is strongly coupled with the Moon's 27 day orbit. As a direct result, from the station's POV, in 24 hours it'll move 13.3 degrees around Earth while Earth would've rotated a full 360 degrees. As far the station is concerned though, it'll observe Earth rotate 360-13.3 degrees or 346 degrees. This means that if the prime meridian were directly below the station, after 24 station hours, the 13th longitude would be beneath it, and after 48 hours the 26th longitude, and so on... For comparison this would place Greenwich at 00:00, ~Naples, Italy at 24:00, Bucharest, Romania at 48:00 and Mecca, Saudi Arabia at 72:00. As you can tell, this would disrupt the proposed equal balance between shifts and world-powers. Unfortunately, this also applies to each other Lagrange point. FURTHER DISCUSSION @ BOTTOM OF POST

Earth-Sun Lagrange Points

Lagrange Points of Earth-Sol System If we look at the Earth-Sun system, many of the same points for the Earth-Moon system still apply. The only major difference here is distance. For Earth-Moon, L1 is 326,390 km as measured from Earth to the Moon. For the Sun-Earth system, L1 is at 57,689,000 km as measured from Earth to the Sun. This would place the satellite well outside the Earth-Moon system.

Sun-Earth L1

For the Sun-Earth L1 point, the 1st argument applies equally; but, the 2nd argument is nearly eliminated. The orbital period of the Earth-Sun Lagrange points is 1 year or 365.25 days. Therefore, during a single 24 hour station day, the station and earth will have moved about 1 degree around the orbit. This comes out to about 1 day for Earth. Technically, after about half a year, at 00:00 hours the other side of earth will be visible; but, now this is minimized dramatically.

EM Radiation

Unfortunately, L1 for the Sun-Earth system experiences a lot of electromagnetic radiation.

Answer: Sun-Earth L2.

If the station were placed at L2, it'd keep a close synchronicity with Earth's rotation without experiencing the EM radiation at L1. I will note that station would be 176x farther from Earth than at the Earth-Moon L2; but, whether or not this is an issue for your world is up to you.


Technically, I assumed Earth's orbit was perfectly circular. In fact this is not true; it's slightly eccentric. As a result, when Earth is closer to the sun during the northern winter months, it travels much faster therefore passing through more radial degrees per day and when Earth is farther from the sun during northern summer months, it travels slower. As a result, my calculations won't be exact; however, given the astronomical scales involved, the error range is small enough to qualify as a back-of-the-envelope calculation.

Discussion on L1 Logistics

Here's a diagram I drew up: L1 Logistics Diagram Fundamentally, because L1 orbits with the moon, for every 24 hour period, points A,B,C,D will always return to the same position; but, the station will orbit 13 degrees to the marked L1 positions, each one above points A, B, C, and D respectively. Now suppose we have three shifts, each representing countries in the Cyan (CY), Magenta (MG), and Yellow (YW) nations below; and, suppose that the shifts are marked as CY from 00:00-08:00, MG from 08:00-16:00, and YW from 16:00-24:00. This means that after 9 days, the Earth will have rotated 9 times, and be at the same position; but, the station will have traveled 119.7 degrees around earth, or ~ 1/3 of its orbit. As a result, when CY begins their shift at 00:00 hundred hours, they'll find that directly below them is the start of MG territory. Normally this would indicate that they are ending their shift and MG's shift is about to begin; but, the station time is 00:00. If you choose L1, your station will need to use a 23.11 H day with shifts every 7.7 Hours (7 hours 42 minutes) to ensure shifts start and terminate above their respective territories.

  • $\begingroup$ What protects the L2 point from radiation, exactly? Is it the magnetosphere of the Earth, or is the L2 still close enough that the Earth physically blocks the view of the sun? $\endgroup$ Commented Sep 19, 2017 at 20:35
  • $\begingroup$ L2 is not protected from solar radiation; but, a lot less solar radiation passes through L2 as composed to L1. It is around 10-100x farther out that the magnetopause and a little beyond Earth's umbra, so Earth's magnetosphere doesn't help and Earth's umbra helps somewhat. The primary difference is that electronics in L2 are not in the direct path of solar radiation as opposed to L1. This dramatically reduces the shielding requirements. Additionally, being farther out helps too (inverse square law). $\endgroup$ Commented Sep 19, 2017 at 21:08
  • $\begingroup$ So if I understand this correctly, then Earth-Moon-L1 would give us some (24h / 360 * 346) 23/24ths of an earthday during a 24 hours period? So by adding another hour (approx 20min to each shift) we could cover a whole earth day - but would eventually shift from earth noon to earth midnight? $\endgroup$
    – dot_Sp0T
    Commented Sep 20, 2017 at 5:30
  • $\begingroup$ @dot_Sp0T I added another section to my OP to answer your question. Essentially, if the station were to be placed at L1, it will require the station to operate in a 23.11 hour day with shifts ever 7.7 hours (7 hours 42min) to ensure shifts start and end above their respective territories. $\endgroup$ Commented Sep 20, 2017 at 15:30
  • $\begingroup$ You sir more than deserve the additional rep :) $\endgroup$
    – dot_Sp0T
    Commented Sep 20, 2017 at 15:43

Given the requirement to view the entire Earth once in a 24 hours period I can see only two solutions.

You can either gain 15 degrees (1 hour) per hour on the rotation of the planet, or lose 15 degrees per hour.

The former requires you to effectively (and exactly) "lap" the planet in a single day of its rotation, meaning you need to move precisely twice as fast. This puts you into a 12 hour orbit around the planet, meaning you make 2 orbits for the Earth's single rotation allowing you to see each point on the Earth by seeing 12 hours (180 degrees) of it on each orbit. Orbital distance 20,200km (approx). This allows the "near Earth" requirement and is the stable, safe option.

The latter forces you into a solar orbit as you have no effective movement relative to the Earth1. You can't just share an orbit though, you need to sit in one of the gravity neutral locations or you'll end up hitting the planet you're sharing orbit with. These safe locations are known as Lagrange points and are numbered L1 through L5 based on location, seen in the image below outlined by gravitational contours.

Wikipedia gravity contour lagrange points

L3 is on the wrong side of the Sun so we'll write that off. L4 and L5 are 60 degrees ahead and behind the planet respectively and they are stable but put you too far away to be useful, they also have a downside in that they're often occupied by trojans. L1, L2 are closer, but unstable, L2 being more stable than L1. This breaks the "near Earth" requirement, but is apparently what you're looking for.

As a side note, NASA advises L5 for long term habitation

1 This is a three body problem, the bodies being the Sun, the Earth, the space-station. The Sun is the datum used to define the relative movement of the other bodies, not the rotation of the Earth.


The title of the question, and the actual question, do not seem to match up.

If different countries are visible during different parts of the day, then the station is not remaining "put". If you want it to "stay put", you need a geostationary orbit, at an altitude of 35,786 km.

If, on the other hand, you don't want it to "stay put", but you want it to pass over the same part of the Earth at the same time every day, that's quite different. In particular, it looks like you want it to make one revolution with respect to the co-rotating frame of the ground once per day, so that it passes over each line of longitude once every 24 hours.

There are two ways to do that: you can approximate it by orbiting really far away- like, as far out as the moon, or farther, so that for all practical purposes the station is completely stationary, and the Earth turns beneath it. If you place it at any of the Earth-Sun Lagrange points, that will work out perfectly- but the closest of those (the L1 point, on a line between the Earth and the Sun) is about 1.5 million km away. Not terribly practical if you need to communicate with the ground in near real time, or if you need to see things clearly.

Probably the better solution is to use a 12-hour sun-synchronous, (aka semi-synchronous) orbit. This way, after one orbit to the same position with respect to the sun (about 2 minutes longer than one sidereal orbit), the Earth will have spun half-way around, and you will have passed over 1/2 of the Earth along the way. And the second orbit, you will pass over the other half of the Earth. Your three shifts will thus be split up over 2 orbits.

The approximate altitude for such an orbit is 20,200 km.

  • $\begingroup$ A sun-synchronous orbit cannot be a semi-synchronous orbit. A sun-synchronous orbit goes roughly perpendicular to the angular rotation. A semi-synchronous orbit goes in the same direction as the rotation. Beyond that, you can't do a sun-synchronous orbit as long as four hours, much less twelve. $\endgroup$
    – Brythan
    Commented Sep 11, 2017 at 23:53
  • $\begingroup$ @Brythan And unfortunate conflict of terminology. I had in mind not an orbit which makes use of precession induced by deviations from perfect sphericality in the Earth to ensure that it aligns with the same longitude at the same solar time every day- but rather, and orbit as measured with respect to the relative position of the sun, as opposed to a sidereal orbit measured with respect to the stars, in this case also producing an orbit which happens to pass over the same point at the same time every day. If you know of more proper terminology for that, please tell, and I'll edit accordingly. $\endgroup$ Commented Sep 12, 2017 at 0:03
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    $\begingroup$ @LoganR.Kearsley the title of the question and the actual question did match up. Your usage of the word put just didn't match mine - call it language barrier. Nevermind though I did my best to amend the title so it doesn't use terminology that you misunderstand. $\endgroup$
    – dot_Sp0T
    Commented Sep 12, 2017 at 9:02
  • $\begingroup$ It looks like the title of the question has been updated to better reflect the question. $\endgroup$
    – user
    Commented Sep 12, 2017 at 11:00
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    $\begingroup$ What about retrograde orbits, you'll get a shorter period so won't have to be so far away. $\endgroup$
    – Innovine
    Commented Sep 14, 2017 at 10:57

Okay, according to this site, one of the formulas for calculating properties of a satellite's orbit is the following:

T^2/R^3 = (4*pi^2)/G*Mcen

T = orbital period in seconds

R = avg Radius in meters

G = gravitational constant

Mcen = mass of the body to be orbited around (in this case Earth)

We know G and Mcen, and we want T to be such that from your point of view on Earth, it takes 24 hours to orbit. We could either orbit slower than Earth rotates, but we'd have to be so far away we're effectively stationary, which has been covered in other answers. The other option is to go faster than earth rotates (specifically twice as fast), thus we'd need to orbit twice in 24 hours which gives us an orbital period of 12 hours.

This is because orbit and space and what-not is all described from a point of view. It's why people thought the sun orbited the earth for a while, before we realised we orbited it and rotated to cause the same effect. From the Moon's point of view it stays still and the earth slowly spins in place in the sky. From Earth's point of view, your stations passes through the sky exactly once a day, at the same time. From some distant observer outside of the solar system the earth spins once every twenty four hours and your station orbits it once every 48. Movement in space is all about perspective.

Thus, inserting all the numbers: (12 hours equals)

43200^2/R^3 = (4*pi^2)/(6.673 x 10-11*5.972 × 10^24)

Multiply G and Mcen:

43200^2/R^3 = (4*pi^2)/3.9851156*10^14

Multiply each side by GMcen and R^3:

43200^2*3.9851156*10^14 = 4*pi^2 * R^3

Multiply out the left side

7.43718*10^23 = 4*pi^2 * R^3

Divide both sides by 4*pi^2

1.88386*10^22 = R^3

cube root everything

26608243.6243 = R

We now have our average orbital radius of 26,608,243.6243 meters. This is from the center of the Earth though. Earth's radius is 6.371 million meters, so subtract that and we get an approximate altitude for your station of 20.2 thousand km.

  • $\begingroup$ I am not sure I follow you. You write "from your point of view on earth" but in the question it is specified that the desired effect of the orbit is for the station to be able to observe a full rotation of earth. $\endgroup$
    – dot_Sp0T
    Commented Sep 15, 2017 at 11:00
  • $\begingroup$ Orbit and space and what-not is all described from a point of view. It's why people thought the sun orbited the earth for a while, before we realised we orbited it and rotated to cause the same effect. From the Moon's point of view it stays still and the earth slowly spins in place in the sky. From Earth's point of view, your stations passes through the sky exactly once a day, at the same time. From some distant observer outside of the solar system the earth spins once every twenty four hours and your station orbits it once every 48. It's all about perspective. $\endgroup$
    – Kyyshak
    Commented Sep 15, 2017 at 11:05
  • $\begingroup$ that's why I put the bounty & notice there. To point out how important a good description is $\endgroup$
    – dot_Sp0T
    Commented Sep 15, 2017 at 11:14
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    $\begingroup$ I've updated my answer to include me explanation at the relevant point. $\endgroup$
    – Kyyshak
    Commented Sep 15, 2017 at 11:17
  • $\begingroup$ "Because the surface itself turns once very 24 hours, we want the orbital period to be 48 hours (or 172,800 seconds)" - there you have error in your assumptions, take look at Separatrix answer $\endgroup$
    – MolbOrg
    Commented Sep 16, 2017 at 16:58

This is a kind of alternative to KareemElashmawy's answer.
As they point out, the best you can do with a single station is to put it in one of the Lagrange points, but these have problems.

The Earth/Sun L1 point is too far away from Earth, and has fairly high radiation.
The Earth/Sun L2 point has lower radiation, but is also far away, and you'll only ever see the night side of earth.
The Earth/Moon L1 point would be close enough, but you'd be tied to the Moon's day/night cycle, which is 28 days long, and so half the cycle you'd see the Earth's day side, and half the cycle you'd see the Earth's night side, which isn't ideal.
None of these points are stable, and so you'd have to constantly monitor your position and apply corrections to stay on station.

Unless you want to use magic level tech you can't have an orbit that stays solely between the Earth and Sun and is within the orbit of the moon. You would essentially have to constantly expend thrust in order to hover in that position.

So I'd like to propose a "PQ" solution; Have three space stations in geostationary orbit. Each station would stay active while the section of the globe that it's monitoring is in daylight. This has a side advantage that there would be no back side of the planet that one of the stations couldn't see which could be useful for emergency and traffic control purposes.


You want your station to observe one full rotation of the earth. I take that to mean that in one day, persons on the station see the entirety of earth pass beneath them.

A geostationary orbit

is a circular orbit 35,786 kilometres (22,236 mi) above the Earth's equator and following the direction of the Earth's rotation.

It helps me to see it. Here is my depiction, with the Earth as a clock face. The station orbits keeping 12 beneath it as the earth turns. To stay above 12 the station must orbit earth in 1 day.

geostationary orbit schematic

Now use that same 24 hour orbital path at 35786 km that the geostationary satellites use, but go retrograde. These are dangerous orbits!

Nevertheless, a satellite in retrograde orbit could pose a major hazard to other satellites, especially if it was placed in the Clarke belt, where geostationary satellites orbit.

Now the satellite moves contrary to the rotation of the earth. At 12 hours the satellite will be again above its starting point, having seen half the entirety of the earth rotate beneath it. Then it sees the other half entire earth move past again and at 24 hours is again above its starting point.

This answers the question in that the station does see one rotation each day. And then another one, the same day. Question does not specify "one and only one" rotation seen per day.

retrograde orbit


Ummm... okay so I think the only solution is to put it in a separate but Earth oriented orbit so either L4 or L5 in a Trojan orbit, or L3 on the far side of the Sun. All these positions allow for uninterpreted observation of the Earth and a stable spacial relationship between the station and the Earth but none of them orbit the Earth. Any orbit of the Earth is going to have an offset period so the station won't be able to operate on an Earth matched clock, the station will go around the Earth in 12 hours while the world turns or some other none observation matched period. L1 and L2 also fit the bill in terms of orbital stability but they are not themselves stable because of Lunar gravity, they are much closer to the Earth though and the cost of propellant to maintain position in the lagrange zone might be worth it to reduce signal delays (light speed signals need 32 minutes for a round trip at L3, something on the order of 11 minutes for the Trojans, 10 seconds for L1 and L2).

Actually a 48 hour equatorial orbit (along the true rotational equator) that chased Earth's orbit should have a "lap time" of 24 hours, I think that fulfills your needs. So orbital radius approximately 67000km, velocity 2.5km/s, aligned with Earth's rotation, someone will probably point out why this doesn't actually work but it works in my head and appears to work on paper. The station goes around the Earth in 48 hours but it's overhead times will match up on a 24 hour cycle.


You mentioned this is a Sci-Fi project, so I'm going to approach this question more as a writer than an astrophysicist. The simplest way I could explain the mechanics would be to initially place the station above the international dateline. Then maintain a geosynchronous orbit opposite to the rotation of the Earth, but only so much so that in 12 hours, the international dateline (because the line circles the globe) will pass again under the station. In 24 hours, the station should return to where it initially was above the international date line.


A slightly different answer than the others seem to be.

Two potential locations for a station to orbit the Earth and be a near match for your requirements.

One is to build the station on the Moon on the side perpetually facing Earth. The other is to set your space station in the same orbit as the Moon, but on the opposite side of the Earth. This will likely take some active station-keeping to maintain the orbit.

For both of these locations, I believe the time it will take for the station to again be directly over the same point on Earth is just under 25 hours (split into three 8 hour, 20 minute Shifts, which includes slight overlap).

This does mean that there will be a gradual shift of which Shift is responsible for which parts of the planet. However, if it is absolutely required that the same people cover the same sections of the planet, it would be fairly simple to set up shift schedules to move personnel from one Shift to another, to match up with their geography.

With redundant personnel in each Shift (required anyway for illness, time-off, vacations to Earth, and random alien abductions), it is absolutely trivial to schedule such a change.


It is not really clear what you exactly mean by "it will stay relatively put in regard to Earth" (I also failed to understand the linked reference).

  • If you want the S.S. to be fixed with respect to Earth, so it will appear to be "suspended" in a fixed point, then you need a geosynchronous orbit. Note a circular geosynchronous will appear to be fixed, a nice variation is to have a highly eccentric geosynchronous orbit that will draw a vertical "figure eight" on earth spending much more time on one than the other; this is used to have a satellite to be "mostly" over USA & Canada while flying over Brazil for a much shorter period.
  • If you want to stay fixed with respect of Earth-Sun axis, so it will appear to pass over a certain point always at the same time of day, then you need to put your station in a Lagrange position of Earth orbit; in that case your best bet is L1 (between Earth and Sun) that is the nearest to Earth, just a bit farther than Moon's orbit; note: L1 is an unstable position, so it needs to be actively kept with attitude engines or similar devices.
  • A reasonable compromise would be to be "fixed" with respect to Earth-Moon system, in this case the L4/L5 positions are advised; You will have a small delay daily in S.S. position with respect to Earth-Sun system, but it will be nearer (same distance as the Moon) and in a stable orbit (no corrections needed).

Please clarify what alternative you need.

  • 1
    $\begingroup$ For clarification on a question you would comment on that question. Answers are for answering - not for commenting. $\endgroup$
    – dot_Sp0T
    Commented Sep 20, 2017 at 5:32
  • $\begingroup$ @dot_Sp0T: I did answer to all three possible scenarios. I don't understand Your comment (and neither the downvote). $\endgroup$
    – ZioByte
    Commented Sep 20, 2017 at 9:03

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